First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightFunctions And Graphschevron_rightExercise 2.1

Solution

(a) $f(x)=x^2-1$

(i) $f(-3)$

\boxed{f(-3)=(-3)^2-1=9-1=8}

(ii) $f(0)$

\boxed{f(0)=(0)^2-1=0-1=-1}

(iii) $f(x-2)$

\begin{aligned} f(x-2)&=(x-2)^2-1\\ &=x^2-4x+4-1\\ &=x^2-4x+3 \end{aligned}

\boxed{f(x-2)=x^2-4x+3}

(iv) $f(x^2+3)$

\begin{aligned} f(x^2+3)&=(x^2+3)^2-1\\ &=x^4+6x^2+9-1\\ &=x^4+6x^2+8 \end{aligned}

\boxed{f(x^2+3)=x^4+6x^2+8}

(b) $f(x)=\sqrt{2x+3}$

(i) $f(-3)$

f(-3)=\sqrt{2(-3)+3}=\sqrt{-6+3}=\sqrt{-3}

So it is not defined in real numbers.

\boxed{\text{Not defined (in real numbers)}}

(ii) $f(0)$

\boxed{f(0)=\sqrt{2(0)+3}=\sqrt{3}}

(iii) $f(x-2)$

\begin{aligned} f(x-2)&=\sqrt{2(x-2)+3}\\ &=\sqrt{2x-4+3}\\ &=\sqrt{2x-1} \end{aligned}

\boxed{f(x-2)=\sqrt{2x-1}}

(iv) $f(x^2+3)$

\begin{aligned} f(x^2+3)&=\sqrt{2(x^2+3)+3}\\ &=\sqrt{2x^2+6+3}\\ &=\sqrt{2x^2+9} \end{aligned}

\boxed{f(x^2+3)=\sqrt{2x^2+9}}